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Consider an infinite-dimensional Hilbert space with an orthonormal basis: $$\{\lvert{n}\rangle\}, n = 0,1,2,...$$ Define a shift operator such that $$S\lvert{n}\rangle = \lvert{n+1}\rangle$$

Then find a representation of $S$ in terms of the above basis and prove that it is not unitary namely: $$S^\dagger S = 1 \ne S S^{\dagger}$$

I tried to consider the representation $$S = \sum_{n} \lvert n+1 \rangle\langle n \lvert$$ $$S^{\dagger} = \sum_{n} \lvert n \rangle\langle n+ 1\lvert$$

But I'm not sure why $S S^{\dagger} \ne 1.$

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  • $\begingroup$ You should be able to compute $SS^\dagger$ explicitly with little effort. What do you get when you try? $\endgroup$
    – J. Murray
    Oct 9 '21 at 18:39
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Look at what happens with $|0\rangle$ when you try to apply $SS^{\dagger}$ to it. The summation in the last two expressions run from $n=0$. As a result, the combination $SS^{\dagger}$ does not contain a term that operates on $|0\rangle$.

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